Предмет: Алгебра, автор: elenashima1977

помогите решить 2 вариант

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Ответы

Автор ответа: oksik1970
0
▪1
а)
 \frac{15 x{y}^{4} }{10 {x}^{3}  {y}^{2} }  =  \frac{3 {y}^{2} }{2 {x}^{2} }  = 1.5 \frac{ {y}^{2} }{ {x}^{2} }
б)
 \frac{ab - b}{ {b}^{2} }  =  \frac{b(a - 1)}{ {b}^{2} }  =  \frac{a - 1}{b}
в)
 \frac{4 {x}^{2} -  {y}^{2}  }{2x - y}  =  \frac{(2x - y)(2x + y)}{2x - y}  = 2x + y
▪2
а)
 \frac{3}{a}  +  \frac{a - 3}{a + 5}  =  \frac{3(a + 5) + a(a - 3)}{a(a + 5)}  =  \frac{3a + 15 +  {a}^{2}  - 3a}{ {a}^{2}+ 5a }   =  \frac{15 +  {a}^{2} }{ {a}^{2} + 5a }
б)
 \frac{2 {x}^{2} }{ {x}^{2} - 4 }  -  \frac{2x}{x + 2}  =  \frac{2 {x}^{2} - 2x(x - 2) }{ {x}^{2} - 4 }  =  \frac{2 {x}^{2}  - 2 {x}^{2}   + 4x}{ {x}^{2} - 4 }  =  \frac{4x}{ {x}^{2}  - 4}
в)
 \frac{7a}{a - b}  - 7 =  \frac{7a - 7(a - b)}{a - b}  =  \frac{7a - 7 a + 7b}{a - b}  =  \frac{7b}{a - b}

▪3
 \frac{5}{ {(a + 2)}^{2} }  -  \frac{5}{ {a}^{2} - 4 }  -  \frac{5}{a + 2}  =  \frac{5(a - 2) - 5(a + 2) - 5(a + 2)(a - 2)}{ {(a + 2)}^{2}  \times (a - 2)}  =  \frac{5a - 10 - 5a - 10 - 5 {a}^{2}  + 20 }{ {(a + 2)}^{2}  \times (a - 2)}  =  \frac{ - 5 {a}^{2} }{({a}^{2}  + 4a + 4)(a - 2) }  =  \frac{ - 5a}{ {a}^{3}  + 4 {a}^{2}  + 4a - 2 {a}^{2}  - 8a - 8}  =  -  \frac{ 5a}{ {a}^{3}  + 2 {a}^{2} -  4a - 8 }

▪4
 \frac{2a - 2c + ax - cx}{ {x}^{2} - 4 }  =  \frac{2(a - c) + x(a - c)}{ {x}^{2} - 4 }  =  \frac{(a - c)(2 + x)}{(x - 2)(x + 2)}  =   \frac{a - c}{x - 2}
... =  \frac{a - c}{x - 2}  =  \frac{6.7 - 5.3}{1.9 - 2}  =  \frac{1.4}{ - 0.1}  =  - 14
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